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91Ó°ÊÓ

Assume that 20 parts are checked each hour and that \(X\) denotes the number of parts in the sample of 20 that require rework. Parts are assumed to be independent with respect to rework. (a) If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that hour 10 is the first sample at which \(X\) exceeds \(1 ?\) (b) If the rework percentage increases to \(4 \%,\) what is the probability that hour 10 is the first sample at which \(X\) exceeds \(1 ?\) (c) If the rework percentage increases to \(4 \%,\) what is the expected number of hours until \(X\) exceeds \(1 ?\)

Short Answer

Expert verified
Perform binomial probability calculations for each part with updated parameters and conditions.

Step by step solution

01

Understanding Part (a)

Firstly, identify the situation as a binomial distribution because we are looking at the number of successes (parts needing rework) in a fixed number of trials (20 parts per hour). Here, the probability of success is \( p = 0.01 \). We want the probability that hour 10 is the first hour where more than 1 part needs rework.
02

Define Success Condition

For hour 10 to be the first hour where \( X > 1 \), none of the first 9 hours can have more than 1 part needing rework. This requires calculating the probability for each hour being less than or equal to 1 part needing rework.
03

Calculate Probability of X ≤ 1 for Binomial Distribution

The probability of at most 1 part needing rework in a single hour is \( P(X \leq 1) = P(X=0) + P(X=1) \), where \( X \sim \text{Binom}(20, 0.01) \).Calculate:- \( P(X=0) = \binom{20}{0} (0.01)^0 (0.99)^{20} = 0.99^{20} \)- \( P(X=1) = \binom{20}{1} (0.01)^1 (0.99)^{19} = 20 \cdot 0.01 \cdot 0.99^{19} \)Add these probabilities to find \( P(X \leq 1) \).
04

Calculate Probability up to Hour 9

Hour 10 being the first with \( X > 1 \) requires that the first 9 hours have \( X \leq 1 \). This is: \( [P(X \leq 1)]^9 \).
05

Calculate Probability for Hour 10

For hour 10, we only need \( X > 1 \). Thus, compute: \( P(X > 1) = 1 - P(X \leq 1) \).
06

Combine Probabilities for Part (a)

The combined probability for hour 10 being the first hour with \( X > 1 \) is:\([P(X \leq 1)]^9 \times P(X > 1)\).Compute this by using values from previous steps.
07

Understanding Part (b)

For part (b), repeat the problem but with increased probability \( p = 0.04 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculations
When we talk about probability calculations in the context of a Binomial Distribution, we're actually exploring the chance that a certain number of events happen, given a fixed number of trials. In this scenario, the trials are the 20 parts checked each hour.By defining the probability of a part needing rework as a 'success,' our job is to figure out the likelihood of this success happening in different scenarios. For example, if the probability of a part needing rework (success) is 1% per part, then we want to calculate this across multiple parts.
We use the Binomial formula:
  • The number of ways to choose k successes out of n trials is given by the binomial coefficient, \( \binom{n}{k} \).
  • The probability of exactly k successes in n trials is then \( \binom{n}{k} p^k (1-p)^{n-k} \), where \( p \) is the probability of success.
This approach helps us determine the likelihood that a specific number of parts will need rework, crucial for making decisions based on statistical analysis.
Independent Events
In probability theory, understanding independent events is crucial for correctly analyzing situations like checking parts for rework. When we say that parts needing rework are independent events, we're stating that the outcome of checking one part doesn't influence the outcome of checking another. This assumption is important because it simplifies our calculations. We don't have to worry about past results affecting future checks. Each check of a part is a separate event, with the same probability of needing rework.
  • An independent event implies no connection or effect between events.
  • This means total probability is a simple product of individual probabilities for each event.
For example:
  • If we check the first part and it needs rework, the probability of any other parts needing rework remains at 1% or 4%, depending on the scenario.
  • Thus, calculating the probability of multiple parts needing rework becomes more straightforward, as each part 'does its own thing.'
Statistical Analysis
Statistical analysis involves examining a situation using statistical methods to understand and interpret data clearly. By employing these techniques, we aim to make informed decisions based on the probability calculations and assumptions about independence. With our exercise, statistical analysis lets us:
  • Determine how likely it is that hour 10 will be the first time more than 1 part fails.
  • Predict the average time it takes before we encounter a specific condition, like exceeding one part that needs rework.
This analysis aims to
  • Detect trends,
  • Understand variability,
  • Anticipate future outcomes,
  • Make decisions based on evidence instead of guesswork.
Through a deeper statistical approach, we not only solve the problem at hand but also enhance our ability to manage systems, improving efficiency and reducing unnecessary work.

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